Consider a system of three identical parallel plate capacitors (numbered 1, 2, 3
ID: 1312592 • Letter: C
Question
Consider a system of three identical parallel plate capacitors (numbered 1, 2, 3) each having a capacitance C = 1 mu F and connected as shown. if 60 V is applied across the terminals of the system and the capacitors become fully charged, how much energy is stored in capacitor #1? [Find the charge on it.] Ans: 8 times 10-4 J. The system is now isolated from the charging source, and a slab of dielectric with k = 2 is slipped between the plates of 21, filling the gap completely. What is the new stored energy in 1? [What remains constant?] Ans: 4 times 10-4 J. What is the new potential difference across the terminals? Ans: 40 V.Explanation / Answer
a) FInding equivalent capacitence of 3 capacitors
2 and 3 are parallel to equivalent of (2+3) = 1+1 = 2 micro farads
Now 1 Micro farad and 2 micro farads are in series
so equivalent capcitence = (2*1)/(2+1) = 0.666 micro farads
So charge across equivalent capacitence = C*V = 0.666 * 60 = 0 Micro coulumbs
Now this charge will be same on 1 micro fards and 2 micro farads ( equivalent o f 2+3)
So, charge on capcitor 1 = 40 Micro coulumbs
Energy stored in capacitor 1 = q^2 / (2*C) = (40^2) *10^-6 / (2* 1) = 800 * 10^-6 J = 8 *10^-4 J
b) Now capacitence = 2* 1 = 2 micro farads
So, energy stored = q^2 / (2*C) = (40^2) *10^-6 / (2* 2) =400 * 10^-6 J = 4 *10^-4 J
c) Potential difference across capacitor 1 = q/C = (40 *10^-6) / (2*10^-6) = 20 V
Potential difference across equivalent capacitence (2+3) = q/C = (40 *10^-6) / (2*10^-6) = 20 V
SO Potential difference across terminals = Potential difference across capacitor 1 + Potential difference across equivalent capacitence (2+3)
Potential difference across terminals = 20 + 20 = 40 V is the answer